I think that this will only happen, though, up until a certain point, because after that the spring will not revert to its original shape3. This point is called the elastic limit. The graph below shows the elastic limit of a spring. As you can see, at a certain point Hooke’s law fails to work4. It starts off as E α L until it reaches the elastic limit where the spring extends more than it would if E α L. This is where the spring starts to stretch out of shape and will not go back to it’s original state.
As my investigation is into the time a spring takes to ‘bounce’ 10 times I will use the information I have just given to explain my hypothesis for the experiment.
My theory is that if a spring extends further then it will take longer to ‘bounce’.
I think that this is because when you add weight to the spring, you give it more potential energy. When you let it go it releases kinetic energy. It will travel further, the more potential energy it has. The spring then needs energy to revert back to its original shape. Because it needs energy it takes more time to get back to the original position. If you add more weight to the spring then it will need more energy to get back. The more weight you add, the more energy is needed and so the slower it takes to get back. This means that if you gave each added weight more pull to start off with, they would probably all take the same amount of time to revert to their original position.
Diagram and List of Apparatus
Apparatus
Method
I will put masses of 100g onto the spring and add 100g each time. I will take a set of 5 results from my experiment. I will then repeat the experiment twice to find repeat readings and averages. I will take readings of how long, in seconds, it takes the spring to oscillate ten times. One oscillation is both an up and a down motion. I will give each a pull of 8cms.
To perform my experiment I will follow these points:
- Set up the apparatus as shown in the diagram.
- Load a weight onto the hanger.
- Use the ruler to measure how far you pull the spring.
- Let go of the spring and start the timer at the same time.
- Count ten bounces and stop the timer.
- Record the result in a table.
- Repeat steps 2-6 adding 1 more weight each time you repeat them.
- Repeat the experiment twice.
I will record my results in these tables:
Fair Test
To make my experiment fair, I will follow these guidelines:
- Make sure you pull the spring down the same amount for each result.
- Make sure you use the same equipment for each experiment, as they could be slightly different.
- Be as accurate as possible. Start the clock as soon as you let go of the spring and stop it as soon as it has oscillated ten times.
- Do not do the experiment near a radiator or anything that generates heat as it may affect the spring, it could change the material properties of the spring.
Safety
To keep my investigation fair I will follow these points:
Follow all normal laboratory rules.
Do not wear open toed shoes as if you drop any weights, open toed shoes do not produce any protection from the weights.
Do not pull the spring down too far as it will cause the hanger to fall off the spring.
Table Of Results
These are the results which I got from my experiment:
Interpretation
As more weight was added the time it took to oscillate increased i.e. 100g took 5.28s to oscillate whereas 500g took 8.81s to oscillate. The weight added is proportional to the time but not directly.
My hypothesis said that the more weight you add to the spring, the longer it would take to oscillate ten times. As this is shown in the results I think my hypothesis is correct.
According to the laws of simple harmonic motion for mass on a spring5, the formula for the period of one oscillation is:
T=2π √m
K
Where
T= Time taken for one oscillation
m= Mass added to the spring
k= The tension required to produce a unit extension (the spring constant)
Using this formula I can now find the formula for 10 oscillations:
10T= 20π √m
k
As we already know the time taken for the oscillations we do not need to use this formula apart from to show why T α m.
The constants in this equation are: 10,20,π,√k. The only variables in the equation are T and √m (√m/√k=√m) as the time changes and so does the
k
mass but everything else stays the same.
As m gets bigger so √m/k gets bigger meaning 20π√m/k gets bigger and so T increases according to the equation.
I pointed out earlier that T α m but not directly. Actually, T α √m.
Evaluation
My results were enough to help me come to the conclusion that when more weight is added, the spring takes more time to revert back to it’s original position.
My results were fairly accurate. They were accurate enough to prove my theory. One of the reasons for this is that we used a digital timer meaning more accuracy.
The result for 300g does not look right if you look at the average results. The results all gradually increase except 300g, which seems to leap. This could be because there was a temperature change or one of the other variables were changed.
I thought that the results would be nearly the same for each experiment but they were not. This is most probably because of numerous reasons:
- The measurements were probably not very accurate as the ruler was just held up to the spring and was measured to the nearest cm.
- The spring and the stop clock were probably not totally in time. The stop clock was most likely started or stopped inaccurately.
I could have made my results better if I had used computer controlled measuring devices, as they are more accurate that human controlled measuring devices.
If I were to do this experiment again I would use computer controlled measuring devices if they were available. I would also have done the experiment to measure load against the extension of the spring to prove what I said in my hypothesis was true.
Appendix
1. Hypothesis
Hooke's Law states that the extension (of say a spring) is directly proportional to the loading force. i.e. if the force is doubled the extension will be doubled, etc.
( search for: “Spring”)
2. Hypothesis
If you hang weights on a spring, it will stretch. The more weight you hang on, the more the spring will stretch.
( search for: “Spring”)
3. Hypothesis
Hooke's Law isn't a 'law' in the sense that it is a statement that is always obeyed. It is rather a statement about how things like wires and springs behave, provided that you don't stretch them too much.
Obviously this can only be true up to a point, as the spring will break eventually. In fact long before it breaks it will cease to obey Hooke's law and won't even revert to its original length when the load is removed.
( search for: “Spring”)
4. Hypothesis
Some materials are elastic. This basically means that they can be stretched when a force is applied to them (like stretching an elastic band or a spring) but when this force is taken away, they return to their normal size.
The reason that rubber bands can do this is because of the arrangement of molecules in the rubber. They are able to slide past each other and so the stretching does not destroy the structure. If you do the same to a material that is not elastic, the molecules are usually torn apart.
However, elastic materials cannot go on being elastic forever. There comes a point when the rubber molecules cannot slide any further and so the band breaks. The force that causes this is called the elastic limit.
In the case of a spring, it is a bit more complicated. A spring is coiled up and will bounce back, until its shape is lost. In either case, the elasticity is destroyed at the elastic limit.
( search for: “Spring”)
5. Interpretation
The extension of a spiral spring which obeys Hooke’s law is directly proportional to the extending tension. A mass m attached to the end of a spring exerts a downward tension mg on it and if it stretches it by an amount l as in the figure below (a), then if k is the tension required to produce unit extension (called the spring constant and measured in N m-1) the stretching tension is also kl and so
mg=kl
Suppose the mass is now pulled down a further distance x below its equilibrium position, the stretching tension action downwards is k (l+x) which is also the tension in the spring acting upwards, this is shown in the picture above (b). Hence the resultant restoring force upwards on the mass
= k (l+x) – mg
= kl + kx – kl (since mg=kl)
= kx
When the mass is released it oscillates up and down. If it has an acceleration a at extension x then by Newton’s second law
-kx = ma
The negative sign indicates that at the instant shown a is upwards while the displacement x is downwards.
k
∴ a= - m x = - ω2x
where ω2 =k/m = a positive constant since k and m are fixed. The motion is therefore simple harmonic about the equilibrium position so long as Hooke’s law is obeyed. The period T is given by 2π /ω, therefore
T= 2π √m
k
It follows that T2 = 4π2 m/k. If the mass m is varied and the corresponding periods T found, a graph of T2 against m is a straight line but it does not pass through the origin as we might expect from the above equation. This is due to the mass of the spring itself being neglected in the above derivation. Its effective mass and a value of g can be found experimentally.
(Advanced Physics Fourth Edition by Tom Duncan. Page 178 Mass on a spring. (a)Period of Oscillations.)
